Problem: Solve for $X$. $X-\left[\begin{array}{rr}1 & -19 & 3 \\ 16 & -4 & 8 \end{array}\right]=\left[\begin{array}{rr}1 & -14 & 3 \\ 12 & -19 & 1\end{array}\right] $ $X=$
Explanation: The Strategy First, we can represent the matrices of the equation with letters, which will make the equation easier to handle. Then we can solve the equation for $X$ and obtain an expression with the letters we defined. Finally, we can substitute back the actual matrices into the resulting expression and simplify it. Solving the equation for $X$ We are given the following equation. $X-\left[\begin{array}{rr}1 & -19 & 3 \\ 16 & -4 & 8 \end{array}\right]=\left[\begin{array}{rr}1 & -14 & 3 \\ 12 & -19 & 1\end{array}\right]$ Let's represent the above matrices as follows. $A=\left[\begin{array}{rr}1 & -19 & 3 \\ 16 & -4 & 8 \end{array}\right] ~~~~~~~~~ B = \left[\begin{array}{rr}1 & -14 & 3 \\ 12 & -19 & 1\end{array}\right]$ Then we can rewrite the equation as follows. $X-A=B$ Now it's simple to solve the equation for $X$. $\begin{aligned}X-A&=B\\\\ X&=A+B \end{aligned}$ Finding $X$ We found that $X=A+B$. Now we can substitute the actual matrices back into the expression and simplify. $\begin{aligned}X&=A+B \\\\&=\left[\begin{array}{rr}1 & -19 & 3 \\ 16 & -4 & 8 \end{array}\right]+\left[\begin{array}{rr}1 & -14 & 3 \\ 12 & -19 & 1\end{array}\right] \\\\\\&=\left[\begin{array}{rr}(1+1) & (-19-14) & (3+3) \\ (16+12) & (-4-19) & (8+1)\end{array}\right] \\\\\\&=\left[\begin{array}{rr}2 & -33 & 6 \\ 28 & -23 & 9\end{array}\right]\end{aligned}$ Summary $X=\left[\begin{array}{rr}2 & -33 & 6 \\ 28 & -23 & 9\end{array}\right]$